9/20
explanation:
7/10 - 1/4
14/20 - 5/20
14-5/20
9/20
Step-by-step explanation:
Hey there!
Given sequences are; 2 , 13 , 24 , 35 , _ , _ .
Now,
Common difference (d) = 2nd term - 1st term. = 13-2 = 11
When we subtract 1st term from 2nd term we find 11 and when we subtract 2nd term from 3rd term we get 11. This means our common difference is 11.
Now, let's find the nrh term of the sequence.
nth term= a1 + (n-1)d ( <em>a1= 1st term, d= common</em> <em>difference</em>)
nth = 2+ (n-1) 11
= 2 + 11n - 11
= 11n - 9
Let's check if we have got nth term correct.
a1= 1*11 - 9 = 2
a2 = 2*11-9 = 13
a3 = 3*11 - 9 = 24
a4 = 4*11-9 = 35
So, we got our nth term.
Let's find remaining sequence.
a5= 5*11 - 9 = 46.
a6= 6*11 - 9 = 57.
Therefore, the remaining terms are : 46 and 57.
<em><u>Hope</u></em><em><u> it</u></em><em><u> helps</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>
Answer:
Correct option: (a) 0.1452
Step-by-step explanation:
The new test designed for detecting TB is being analysed.
Denote the events as follows:
<em>D</em> = a person has the disease
<em>X</em> = the test is positive.
The information provided is:

Compute the probability that a person does not have the disease as follows:

The probability of a person not having the disease is 0.12.
Compute the probability that a randomly selected person is tested negative but does have the disease as follows:
![P(X^{c}\cap D)=P(X^{c}|D)P(D)\\=[1-P(X|D)]\times P(D)\\=[1-0.97]\times 0.88\\=0.03\times 0.88\\=0.0264](https://tex.z-dn.net/?f=P%28X%5E%7Bc%7D%5Ccap%20D%29%3DP%28X%5E%7Bc%7D%7CD%29P%28D%29%5C%5C%3D%5B1-P%28X%7CD%29%5D%5Ctimes%20P%28D%29%5C%5C%3D%5B1-0.97%5D%5Ctimes%200.88%5C%5C%3D0.03%5Ctimes%200.88%5C%5C%3D0.0264)
Compute the probability that a randomly selected person is tested negative but does not have the disease as follows:
![P(X^{c}\cap D^{c})=P(X^{c}|D^{c})P(D^{c})\\=[1-P(X|D)]\times{1- P(D)]\\=0.99\times 0.12\\=0.1188](https://tex.z-dn.net/?f=P%28X%5E%7Bc%7D%5Ccap%20D%5E%7Bc%7D%29%3DP%28X%5E%7Bc%7D%7CD%5E%7Bc%7D%29P%28D%5E%7Bc%7D%29%5C%5C%3D%5B1-P%28X%7CD%29%5D%5Ctimes%7B1-%20P%28D%29%5D%5C%5C%3D0.99%5Ctimes%200.12%5C%5C%3D0.1188)
Compute the probability that a randomly selected person is tested negative as follows:


Thus, the probability of the test indicating that the person does not have the disease is 0.1452.